Find the smallest number which when divides 28 and 32, leaving remainders 8 and 12 respectively.

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#### Solution

Let the required number be x.

Using Euclid’s lemma,

x = 28p + 8 and x = 32q + 12, where p and q are the quotients

⇒28p + 8 = 32q + 12

⇒28p = 32q + 4

⇒7p = 8q + 1….. (1)

Here p = 8n – 1 and q = 7n – 1 satisfies (1), where n is a natural number

On putting n = 1, we get

p = 8 – 1 = 7 and q = 7 – 1 = 6

Thus, x = 28p + 8

= 28 × 7 + 8

= 204

Hence, the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 is 204.

Concept: Euclid’s Division Lemma

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